A friend of mine posed a problem on a mathematics discord server.

The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.

I immediately thought of employing the binomial theorem, as what was required, actually. But, I decided to do another method, which led me to the question of how we can solve such problems without using the binomial theorem.

How can we solve this question without the binomial theorem or newton's method?

My Attempt:

Lemma: $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc.$$

Note that $$(2+px)^6=(2+px)^2(2+px)^2(2+px)^2$$ $$=((px)^2+4px+4)((px)^2+4px+4)((px)^2+4px+4)$$ Therefore, in the lemma, substitute $x\mapsto (px)^2$ and $a,b,c\mapsto 4px+4$. It follows $$(2+px)^6=(px)^6+12(px+1)(px)^4+48(px+1)^2(px)^2+64(px+1)^3.$$ We can ignore the first two terms since they contain no strict $x^2$ coefficient. Thus, upon expanding $$48(px+1)^2(px)^2+64(px+1)^3$$ it follows the coefficient of $x^2$ is $240p^2$. $$240p^2=60\tag{$p>0$}$$ $\therefore p=\frac 12$.

  • 2
    $\begingroup$ You write The coefficient of the expansion of $(2+px)^6$ is $60$. Do you mean The coefficient of the $x^2$ term in the expansion...? $\endgroup$ Jun 4 '20 at 4:01
  • $\begingroup$ @AndrewChin yes, that's what I meant. My apologies - that lack of specificity presumably stumped many users, lmao. $\endgroup$
    – Mr Pie
    Jun 4 '20 at 9:23
  • $\begingroup$ You are basically explicitly expanding $(a+b)^6$, though you are doing it in two steps - first the cube $(a+b)^3$ and then substituting $(a^2, 2ab+b^2)$ for $(a,b)$ to square it. Of course you are also leaving out any terms you don't need along the way. $\endgroup$ Jun 4 '20 at 9:46
  • $\begingroup$ @JaapScherphuis Eh, you are right, so imma just delete this question altogether. I mean, I don't suppose it contributes to anything, and perhaps I merely overreacted at the somewhat unorthodox method with which I approached this problem. $\endgroup$
    – Mr Pie
    Jun 4 '20 at 10:16

This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...

Let $$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$

Differentiate twice:

$$\begin{align} p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2) \\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x) \end{align}$$

Evaluate at $x=0$:

$$30p^2\cdot 2^4=120\Rightarrow p^2=\frac{1}{4}\Rightarrow p\underset{p>0}{=}+\sqrt{\frac14}=\frac12.$$

  • $\begingroup$ That is very neat! (You may have made a typo at the last part, but I understand what you mean :P) It doesn't directly use the binomial theorem but I never thought about differentiating to reduce the degree of $\mathcal O$. Thank you! You answered just in time too $-$ I was ready to delete my question, lol. $\endgroup$
    – Mr Pie
    Jun 4 '20 at 10:36
  • $\begingroup$ @MrPie it depends on how you prove that the derivative of $x^n$ is $nx^{n-1}$... I suppose it could be done with induction and so this is independent of the binomial theorem. $\endgroup$ Jun 4 '20 at 12:49
  • $\begingroup$ I'd just go with limits by first principles to prove that, upon differentiation, $x^n$ does indeed map to $nx^{n-1}$. $\endgroup$
    – Mr Pie
    Jun 4 '20 at 23:11
  • $\begingroup$ @MrPie how do you do that not using induction nor the binomial theorem? $\endgroup$ Jun 5 '20 at 8:05
  • 1
    $\begingroup$ @mrpie I suppose you could argue that $(x+h)^n=x^n+nhx^{n-1}+\mathcal{O}(h^2)$ without the Binomial Theorem but I would just use induction. $\endgroup$ Jun 6 '20 at 7:48

Here are three different methods. The first one is recommended, the others are just for fun and curiosity.

Combinatorial approach:

We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. Here we are looking for a positive solution $p$ of \begin{align*} 60=[x^2](2+px)^6\tag{1} \end{align*} Multiplying $(2+px)^6$ out we get a contribution to $x^2$ if and only if we select from two factors $2+px$ the term $px$ giving us $p^2x^2$ and taking from the other $4$ factors $2$ giving us $2^4=16$. Since we can select two factors from $(2+px)^6$ in $\binom{6}{2}$ ways, we obtain \begin{align*} \color{blue}{60}=[x^2](2+px)^6=\binom{6}{2}p^22^4=15p^2\cdot 16\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Algebraic approach:

A somewhat cumbersome but simple approach is to iteratively work through the linear factors $2+px$.

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=[x^2](2+px)(2+px)^5\\ &=\left(2[x^2]+p[x^1]\right)(2+px)(2+px)^4\tag{1}\\ &=\left(4[x^2]+4p[x^1]+p^2[x^0]\right)(2+px)(2+px)^3\tag{2}\\ &=\left(8[x^2]+12p[x^1]+6p^2[x^0]\right)(2+px)(2+px)^2\\ &=\left(16[x^2]+32p[x^1]+24p^2[x^0]\right)(2+px)(2+px)^1\\ &=\left(32[x^2]+80p[x^1]+80p^2[x^0]\right)(2+px)\\ &=160p^2+80p^2\tag{3}\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.


  • In (1) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

  • In (2) and the following lines we determine the coefficients according to the rules from (1), i.e. \begin{align*} &\left(a[x^2]+b[x^1]+c[x^0]\right)(2+px)=2a[x^2]+(2b+ap)[x^1]+(2c+bp)[x^0] \end{align*}

  • In (3) we select the coefficients accordingly.

Complex analytic approach:

This variant shouldn't be taken too serious. It's just for fun and in fact based on the first method. We recall the residue theorem which tells us that integrating along a circle with radius one around the origin we have \begin{align*} [x^2](2+px)^6=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx \end{align*}

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx\\ &=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{\left(2+pe^{it}\right)^6}{e^{3it}}\,ie^{it}\,dt\tag{2}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\left(2+pe^{it}\right)^6e^{-2it}\,dt\tag{3}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\binom{6}{2}2^4p^2\,dt\tag{4}\\ &=\frac{1}{2\pi}\binom{6}{2}2^4p^2\int_{0}^{2\pi}\,dt\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.


  • In (2) we use the substitution $x=e^{it}, dx=ie^{it}dt$.

  • In (3) we note $\int_{0}^{2\pi}e^{-kit}\,dt =0, k\in \mathbb{Z}$ which is due to Euler's identity $e^{2\pi i}=1$. So, everything vanishes in (3) besides the constant term.

  • In (4) we select the constant term in fact with the same considerations we used in the first approach.

Note: Another approach might be using shifted series multisection in order to filter the wanted coefficient of $x^2$.

  • $\begingroup$ Wow, that is really efficient! I have never seen this method before, so thank you very much for showing this to me. It appears it can be widely applicable. +1 :) $\endgroup$
    – Mr Pie
    Jun 6 '20 at 5:32
  • 1
    $\begingroup$ @MrPie: You're welcome. Another application is e.g. when proving binomial identities. $\endgroup$
    – epi163sqrt
    Jun 6 '20 at 5:51
  • $\begingroup$ @MrPie: I've extended the answer somewhat. $\endgroup$
    – epi163sqrt
    Jun 6 '20 at 11:33
  • $\begingroup$ That complex analytic approach.... I am gonna need a whole book and a half to grasp that XD $\endgroup$
    – Mr Pie
    Jun 7 '20 at 4:52

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